106,269 reputation
696194
bio website von-eitzen.de/math/tntrep.xml
location Bonn, Germany
age 48
visits member for 1 year, 10 months
seen 45 mins ago

I did study math and had a knack for it, but I am sooo out of that business now ...


Sep
10
answered Proving that 4 specified sets are not algebraic
Sep
10
answered Calculating fluid pressure for ruptured eardrum
Sep
10
answered Can we have $\displaystyle \lim_{x \to f(x)}$?
Sep
10
comment Calculating a rampdown interval
What constitutes a "good" formula? What's wrong with slightly less than two hours between any two attempts?
Sep
10
comment Find all $n\in\mathbb{Z}^+$ satisfying : $n\mid(2^{\varphi(n)}+3^{\varphi(n)}+\cdots +n^{\varphi(n)}).$
You forgot $n=1$, but otherwise you just beat me with the $r=4$ case
Sep
10
awarded  Nice Answer
Sep
10
comment How to prove a function is periodic?
I assume one should add "for some $n\in\mathbb Z$" somewhere.
Sep
10
answered How to prove a function is periodic?
Sep
10
revised How to prove a function is periodic?
LaTeX markup
Sep
10
comment How can one prove that $e<\pi$?
I always thoght that comparing $e^\pi$ with $\pi^e$ (without calculator) was the only funny specimen of "this kind of problems".
Sep
10
answered How can one prove that $e<\pi$?
Sep
10
answered elements with bounded conjugate in a number field
Sep
10
comment Divisibility by 7
@DonAntonio Looks wrong at first glance because the difference between $10x+y$ and $x-2y$ is $9x+3y$ and has no reason to be a multiple of $7$. But indeed the trick makes use simply of $7|21$, i.e. that $10x+y-10\cdot(x-2y)=21y$
Sep
10
comment Prove that the set of all algebraic numbers is countable
Your proof is fine, and you don't even need the fundamental theorem of algebra: It is sufficient to know that there are at most $k$ distinct roots.
Sep
10
answered Limit does not exist
Sep
10
awarded  discrete-mathematics
Sep
9
comment Cardinality and bijections
And a possible hint for the first one is: continued fractions
Sep
9
answered Dimension of subspace
Sep
9
comment Proving that $A\subseteq f^{-1}(f(A))$ and that if $f$ is injective $A=f^{-1}(f(A))$.
At this basic level one has to ask: Is it already known that $X\subseteq Y\subseteq S$ implies $f(X)\subseteq f(Y)$ and that $X\subseteq Y\subseteq T$ implies $f^{-1}(X)\subseteq f^{-1}(Y)$?
Sep
9
answered Techniques of Proof